Solve the following simultanious equations: zy=28 and 2z-3y=13

zy=28 so y=28/z13=2z-3y13= 2z - (28 x 3)/z13=2z-84/zmultiply each side by z to give 13z = 2z2-84rearange for a quadratic2z2-13z-84=0solve by factorising(2z+8)(z-21/2)z= -4 0r 21/2substitute -4 into both initial equations-4y=28 and -8-3y=13 both give y=-7substitute 21/2 into both inital equations21y/2=28 and 21-3y=13 gives y=8/3(-4,-7) and (21/2,8/3)

OB
Answered by Oliver B. Further Mathematics tutor

2267 Views

See similar Further Mathematics GCSE tutors

Related Further Mathematics GCSE answers

All answers ▸

Lengths of two sides of the triangle and the angle between them are known. Find the length of the third side and the area of the triangle.


Simplify fully the expression ( 7x^2 + 14x ) / ( 2x + 4 )


Find any stationary points in the function f(x) = 3x^2 + 2x


Find the gradient of the line x^2 + 3x - 6 at the point (5,34)


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning